Question

A car’s stopping distance in feet is modeled by the equation d(v)= 2.15v^2/58.4f where v is the initial velocity of the car in miles per hour and f is a constant related to friction. If the initial velocity of the car is 47 mph and f = 0.34, what is the approximate stopping distance of the car?

a. 21 feet

b. 21 miles

c. 239 feet

d. 239 miles

Answer 1

239 pts (Letter C)

Explanation:

Got it correct

Answer 2

Answer: C is the correct answer.The approximate stopping distance of the car =239 feet.

Explanation:

Given: A car’s stopping distance in feet is modeled by the equation

`d(v)=2.15(v^2)/(58.4f)` where v is the initial velocity of the car in miles per hour and f is a constant related to friction.

If the initial velocity of the car v= 47 mph and f = 0.34 then

The distance of the car
`d(47)=2.15\frac{{47}^2}{58.4*0.34}=2.15(2209)/(19.856)=2.15*111.25=239.18`

≈239 feet [ round to nearest ones]

Therefore ,the approximate stopping distance of the car =239 feet.